Turbulence Part1

8/9/2019 Turbulence Part1

1/39

Turbulence

T U R B U L E N C E

Lecture Note

2009 Fall

Parallel Computing Lab.

Hanyang Univ.

http://vortex.hanyang.ac.kr

Parallel Computing Lab. Hanyang Univ. 1


2/39

Turbulence

.

A random variable u is defined by :

A function which assigns a value to every outcome of the experiment)(u

Random variable u must have the same probability of generating a given outcome -> identically distributed

An ensemble average is defined by :(true avera e)

=

=N

i

i

N

uN

u1

1lim

But, in reality we never have an infinite # of siu

can never compute ensemble average

We can define an estimator for the average based on a finite # of siu

=N

iNu

Nu

1 itself a r.v.

=

Questions : Is this estimator unbiased?

(Dose it converge to correct answer?)

Does it converge at all?



3/39

Turbulence

It is im ortant to be able to tell how a r.v. is distributed about the mean (ensemble avera e)

Define : Variance of r.v. u

{ } ( )22var uuu u

: standard deviation of uu

variance is also called 2nd central moment

Suppose 2 r.v.s are identically distributed, these must have the same variance

an e ne g er moments

nth moment : =

N

j

n

jN

nu

Nu

1

1lim

n=1 : mean

n=2 : mean square

central nth moment (first substract mean value)

=

Nn

jN

n uuN

uu1

)(1

lim)(

How does relate to ?{ }uvar 2u

Splitting u into mean and fluctuation partuuuu +=

mean fluc.

[ ]22 )( uuuu +=

22



4/39

Turbulence

cf. Eas to see from def. of ensemble ave. that arithmatic o erations and avera in o arations commute

e.g. =

=

+=+

N

i

iN

N

i

iN

vN

uN

vu11

1lim

1lim

==N1

=

i

iiN N 1

222 )()(2 uuuuuuu ++=

{ }uuu var022 ++=

(mean square) = (square of man) + variance

or { } 22var uuu =

same mean , different variance

same mean , same variance



5/39

Turbulence

Can we characterize the am litude distribution of si nal?

- make a frequency of occurance diagramConsider N samples

How many sample fall in a particular window

Let # of realizations increase while window size remain same

eve


u


6/39

Turbulence

= ,

0),(lim0

=

uuHu

But : Probability density function)(),(

lim0

uBu

uuH

u=

p. .

a double limit

N

0u

m t ng curve o stogram

Properties of PDF

B(u) 0

Prob {c u c+dc} B(c)dc- follows from def. of B(c) from histogram in limit as ,0c N

B(c)


dcc

c


7/39

Turbulence

c

)()(}{ 1221 cFcFcucp =

)()( cFdc

dcB = F(c)

1)( =

dccB

,1)( =F 0)( =F-

B(c)

c

Ex) 1. Drive pdf for a sine wave.

2. Drive pdf for a triangle wave.

3. Drive pdf for random square wave.

4. What happens to 1. 2. 3. if frequency of signal is changed?

Evaluation of moments from pdf

Ensemble average :

==

N

i

iN

u

N

u1

1lim uu

Nii ++=

)(

1lim

1111

= duuuB )(


pro a y average an ensem e average are e same


8/39

Turbulence

th

= duuBuu nn )(

nth central moment

= duuBuuuu nn )()()(

variance var{u} = 0)()(

duuBuu n (cf. )0)( uB

third central moment =

duuBuu )()( 3

)]()([2

1)]()([

2

1)( uBuBuBuBuB ++=

even odd

- variance can be zero only when (steady signal or all values exactly same))()( uuB =

- t r centra moment s zero u s even, can e non-zero on y u as an o part.

- As a measure of symmetry of pdf

3)( uu 2

3

}][var{u nondimensionalize



9/39

Turbulence

Contribution of isbigger when

3)( uu 0


10/39

Turbulence

Bivariate random variablesthe joint statistics how they interact with each other

v

o n momen s

Consider two r.v.`s u & v

2u 2v, - single

2)( uu - variance

2uv vu

2,joint ),( 11 vu

u

))(( vvuu (cross - ) correlation

(cross - ) covariance

u

1u

v

1v



11/39

Turbulence

),,,(lim),(

00

vuvuHvuB

vu

N=

Properties of JPDF

0),( vuB

1),( =

dudvvuB

),(},{ 11 vuFvvuup = : joint prob. distribution function

vu

vuvuB

=

,),(

),( = uFFu

vFF =

marginal PDF = single variable P.D.F

v

== uF

dvvuBuBu

u ),()(

==v

FduvuBvB v

v ),()(



12/39

Turbulence

),( vuB

u1u

1v

Suppose we want the statistics one var. given, a particular value of the other- conditional prob. Say the particular value of v is v1

( profiles )

/(,( vvuBvuB ==

=

== duuuBvudvBudududvvuuBu u )(),(),(



13/39

Turbulence

m-nth joint moment

dudvvuBvuvu nmnm ),(

=

m-nth joint central moment

dudvvuBvvuuvvuu nmnm ),()()()()( =

for m=1, n=1

=== dudvvuBvvuuvuvvuuCuv ),())(())((

=

is equivalent to the product of inertiathus, a measure of asymmetry of B(u, v)

then we say & (or & ) are uncorrelateduv

vu

Note vuuvvuvuvvuuuv +++=++= ))((0 0

uncorrelated : vuuv =

If two r.v.`s have a non-zero mean

yield the product of their averages,even if they are uncorrelated.

Therefore the product of mean values has`


no ng o o w w e er or no e r.v. sare correlated, only fluctuation part is of interested


14/39

Turbulence

Define (cross-)correlation coefficient

vu

uvvu

vuvu

=

}var{}var{

if u & v perfectly correlated : 1uv

uncorrelated :

so must always 1uv

0uv

perfectly anti-correlated : 1=uv



15/39

Turbulence

Statistical Inde endence

: occurrence of u in no way affects the probability of occurrence of v, and conversely

)()(),( vBuBvuB vu=

JPDF = product of MPDFs

Questions

1. If two r.v.s are statistically independent, are they always uncorrelated? yes

0((((((((

))()(,())((

===

=

dvvBvvduuBuududvvvuuvBuB

dudvvvuuvuBvvuupf.)

=0 =0

2. If two r.v.s are uncorrelation, are they necessarily statististical independence?

Assume two r.v.s & with . These are clearly not statistical independence0vu

No combination of these two r.v.s can be statistical independence.

However, we can find a combination which has zero correlation.

u v

22

22))((

,

,

vvuvuuvuvuyx

vuyvux

vuyvux

=

+=+=

=+=

=+=e.g.)



16/39

Turbulence

22 =& are not statistical inde endence since & are not. If then and=u vx y

there is no correlation in spite of statistical independence.

uncorrelated Statistical independence Uncorrelated

statistical independence

c . var ate orma auss an str ut on

Suppose u & v are normally distributed r.v.s with standard deviation given by & ,

and with correlation coeff.u v

vuuv

vvuu

))((

=

2)( uu

Either r.v. taken separately has a Gaussian marginal distribution.

22

2)( u

u

u euB

=e.g.)

Bivariate Normal P.D.F

( )

+

=

2

2

2

2

2

)())((2)(

12

1exp

12

1),(

vvuuuvuvvu

vvvvuuuuvuB

cf. Central limit theorem (Law of large numbers)



17/39

Turbulence

Estimation of statistical ro erties from a finite number of realization

- never have infinite number of realizations- so how good are our estimatiors based on a finite number

uuN N

Nu

Question : Does it converge to right answer?

i.e. as

taking the average of the estimator, e.g.

N

M

k

NM

uuM k

==

1

1lim or

NNuNNuduBuu

N

= )(

sampling P.D.F

Bias : Estimator is unbiased if the average of estimator yields the true averageno systematic errorconverge to correct value

uuNN

uN

uN

u ii

iN ==== =

111

1

This estimator is unbiased

N

{ } ( )2var NNN uuu

Does { } ?as0var NuN



18/39

Turbulence

Define variability of estimator :

{ }22 var

uuN

if 0 as N, then estimator converges to true valuecan show that

2

2 }var{1

u

u

N=

i.e. variability of estimator is equal to variability of r.v. itself divided by the number of independentrea zat ons

error :uN

u1

=

N

Nu

NuB



19/39

Turbulence

How many flips are required to measure expected value of to within 1% error?

Ex) Coin flip experiment

4

1)(}var{

2

2 ==

=

uuu

u

since

= 1

0

u

4

24

2

2

42

2

104

111

1001.0}var{

==

=

==

N

u

u

u

N

2

1

u

How well can we do with 100 flips?

%101.01100

11====

uN

u



20/39

Turbulence

Stationar Random Process (stochastic Process)

u(t)

t

For stationary random process, its PDF and moments are time-independent (independent of the origin of time).

This will only approximate a real process, since stationary random process must go on forever.



21/39

Turbulence

Ensemble average

dttuT

uT

T = 0 )(1

lim

can define the estimator (time average)

dttuT

uT

T = 0 )(1

itself random

dttudttuuTT

T == )(1)(1

for stationary random process

constutu ==)(

TT uudtu

T

u ==0

is unbiased estimator

Does as ? ( as ?)0}var{ tu T uuT T

))((1

)(1

)()(}var{

2

0

2

0

22

dtutu

T

udttu

T

uuuuu

TT

T

TTTT

=

=

==

)()'(')('])'(][)([

'])'(][)([1

0 02

Ctutuutuutu

dtdtutuutuT

T T

==

= : autocorrelation


tt= 'where


22/39

Turbulence

Why is a function of only?

This can not depend on time itself (i.e. t) since the process is assumed stationary.

)'(')(' tutu tt= '

Why does as ?0)( C

u(t)c()

t

and becomes uncorrelated as

)(' tu )(' +tu

.

- they have finite memories They become uncorrelated with themselves.



23/39

Turbulence

What is a reasonable measure of how long a process is correlated with itself?

Define integral scale as a measure of the memory of process

c()

2/

I

( )

( ) constuuc

dcc

===

=

2/

0

}var{0

)(0 I

for stationary

2/

// )()(

)0(

)()(

u

tutu

c

c

+==

= )( dI


0


24/39

Turbulence

Not all process have integral scale

if then

=0 0)( d

()

0=I

2effI

eff =01

or

Back to ,}var{ Tu =T T

T dtdtttcT

u0 0

//

2)(

1}var{

=T T

dtdtttT

u

0 0

//

2)(

}var{

After a partial integration ,

du

u

T

T

= 1)(}var{2

}var{

Since as)(1)(

T

T

}var{2

)(}var{2

}var{ udu

uT

T

I= (for T >> )I



25/39

Turbulence

2 }var{2)var( uu cf. variability 22

uTu

Therefore asuuT T

Hence for stationary random processes we dont need to perform many experiments to determine statistics.

Compare2

2 }var{1

u

u

N= : N independent realization

2

2 }var{2

u

u

N

= : integral scale, T record timeI

Obviously the effective number of independent realization is

I2

TNeff =

The segments of our time record of two integral scales in length contribute to the average

as if they were statistically independent.

u(t)

t2I


T b l


26/39

Turbulence

2 )(')(' rxuxu +x

DU

0.1xl 0 04x Spatial integral scale

l

cU

l=I where = local convection velocitycU

l 0.04x

EEc . .

Suppose we measure at x/D = 3

04.0 =xD .6.0 DUE

How long must we measure to obtain mean velocity to within 1%

01.199.0

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Turbulence Part1

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